# Thread: Determine the Equation Of Sinusodial Function

1. ## Determine the Equation Of Sinusodial Function

Could someone please explain how I would find the ht of sine and cosine on this graph? I don't understand what the difference is. Also how would I find the period? Is it the same for cosine and sine? Thank-you to anyone that could help!! Not in radian form please...

2. ## Finding a sine and cosine function for a given graph

Hello mmhaaz
Originally Posted by mmhaaz
Could someone please explain how I would find the ht of sine and cosine on this graph? I don't understand what the difference is. Also how would I find the period? Is it the same for cosine and sine? Thank-you to anyone that could help!! Not in radian form please...
First: the difference between the graphs of sine and cosine: there is none, except a 'phase shift' of $90^o$. In other words, if you shift the cosine graph $90^o$ to the right, you get the sine graph. See the attached diagram.

Note in particular the 'starting' values; i.e. the values when $x = 0$:

• $\cos 0^o=1$
• $\sin0^o=0$

Then, the particular questions relating to the graph you've attached.

The scale on the vertical axis isn't very clear, but it looks as if one square = one unit.

So, on that assumption, the graph has a maximum value of $1$ and a minimum value of $-5$. The sinusoidal axis is the horizontal line that divides the graph exactly in half. So this is the line $y = -2$ (which I think is what you got).

Then, the amplitude is the vertical distance from the sinusoidal axis to one of the peaks or troughs. So that's $3$ (again, I think you got that).

From these two facts, we can deduce that the equation of the graph will be of the form:
$y=-2+3\sin$(something) or $y = -2+3\cos$(something)
The question then refers to the 'starting point', based on a sine and then a cosine graph. I assume that means the point which corresponds to the point where $x = 0$ on the graphs of the basic graphs (the ones in my diagram).

So, for sine, this is the point that corresponds to $(0,0)$ on the basic graph - in other words one of the points where the graph crosses the sinusoidal axis and is moving upwards in a direction from left to right. There's more than one such point, but the obvious one is $(30, -2)$. (Can you see that another one will be $(210,-2)$?)

For the cosine function, the starting point will be the point corresponding to $(0,1)$ on the basic graph. So we're looking for a peak (maximum) value: the obvious one here is $(75,1)$.

The period is the horizontal distance between two consecutive corresponding points on the graph. We said above that two possible starting points for the sine graph are $(30,-2)$ and $(210,-2)$. So the period is the distance between these two points. That's $180^o$.

Divide by $360^o$ for the next answer: $\frac{180}{360}=\frac12$.

If we divide $x$ by this fraction ( $\tfrac12$) we get $2x$, which gives us the coefficient of $x$ in the functions we're looking for. (A complete period for $y = \sin x$ is when $x$ goes from $0^o$ to $360^o$; so if we want a period of $180^o$, we shall need values of $2x$, because they will go from $0^o$ to $360^o$ when $x$ goes from $0^o$ to $180^o$.)

The 'starting points' that we've worked out above tell us the angle that has to be added or subtracted from $x$ to give the correct starting value - the phase-shift.

So, for the sine function we want to start at $x = 30$, so the function is
$y=-2+3\sin(2[x-30])$
And for the cosine function, we want to start at $x = 75$, so the function is
$y=-2+3\cos(2[x-75])$